MATHEMATICS HIGH SCHOOL

HELP NEEDED, GIVING BRAINLIEST!!Which of the following are the vertices of the image of the figure below under the translation (x , y) ---> (x + 4, y - 2)? SELECT ALL THAT APPLY

A. (- 1, 2)
B. (- 9, 6)
C. (- 2, 7)
D. (6, 3)
E. (1, - 4)
F. (- 7, 0)
HELP NEEDED, GIVING BRAINLIEST!! Which of the following are the - 1

Answers

Answer 1
Answer:

Answer:

d b f

Step-by-step explanation:

Related Questions

Determine the slope of the line that passes through each pair of points.1.(5,1) and (2,7) 2.(5,3)and(-2,3)
A principal of $835 in an account at 3% per quarter simple interest
Evaluate the expression 36y + 9 if y =1/6
Which off the following expressions is equal to -3x^2-12A. (-3x+2i)(x-6i)B. (-3x-6i)(x+2i)C. (-3x-6i)(x+2i)D. (-3x-4i)(x-3i)
Which fraction represents the decimal 0.12 a) 1/12 b) 3/25 c) 4/33 d ) 33/4

In a pot worth $2.35, there are 6 quarters, 5 dimes, 5 pennies, and the rest of the coins are nickels. What is the ratio of nickels to dimes?

Answers

Answer:

6:5

Step-by-step explanation:

It is given that a pot worth $2.35 and there are 6 quarters, 5 dimes, 5 pennies, the rest of the coins are nickels.

We know that

$1 = 100 cents

1 penny = 1 cent = $0.01

1 nickel = 5 cents. = $0.05

1 dime = 10 cents. = $0.10

1 quarter = 25 cents = $0.25

The value of 6 quarters is

6* 0.25=1.50

The value of 5 dimes is

5* 0.10=0.50

The value of 5 pennies is

5* 0.01=0.05

Let x be the number of nickels. So, the value of x nickels is

x* 0.05=0.05x

Total value of 6 quarters, 5 dimes, 5 pennies, and x nickels is

Total =1.50+0.50+0.05+0.05x

Total =2.05+0.05x

It is given that the pot worth is $2.35.

2.05+0.05x=2.35

Subtract 2.05 from both sides.

0.05x=0.30

Divide both sides by 0.05.

x=6

The number of nickels is 5.

(Nickel)/(Dimes)=(6)/(5)=6:5

Therefore, the ratio of nickels to dimes is 6:5.

In a pot worth $2.35 containing 6 quarters, 5 dimes, 5 pennies, and some nickels, the ratio of nickels to dimes is 6:5.

To find the ratio of nickels to dimes, we need to determine the number of nickels and dimes in the pot. We know that there are 6 quarters, 5 dimes, and 5 pennies in the pot, which is a total of 16 coins. Therefore, the number of nickels should be the difference between the total number of coins and the sum of quarters, dimes, and pennies.

The total value of the coins in the pot is $2.35. Since 6 quarters are worth $1.50, 5 dimes are worth $0.50, and 5 pennies are worth $0.05, the remaining value should come from the nickels.

Thus, the value of the nickels is $2.35 - $1.50 - $0.50 - $0.05 = $0.30. Since each nickel is worth $0.05, the number of nickels is $0.30 ÷ $0.05 = 6.

The ratio of nickels to dimes is therefore 6:5, which means that for every 6 nickels, there are 5 dimes.

Learn more about ratios here:

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Which number is a perfect cube?
09
18
27
O 36

Answers

Answer:

36

Step-by-step explanation:

a cube has 6 sides

6 squares on each side

36 divided by 6 = 6

6 x 6 = 36

hope this helps!

Find the simplified form of (t^4/10y^4)^-4

Answers

The answer is [(10,000*y^16)/t^16]. At first, you have to distribute the exponent -4 to the numerator and denominator by virtue of the law of exponent, (a/b)^n = [(a^n)/(b^n)]. After distribution, you get (t^-16)/[(10^-4)*(y^-16)]. To simplify the negative exponent, follow the law of exponent: a^-m = 1/a^m. Thus, you now have [(10^4)*(y^16)]/(t^16). Hence, the answer is [(10,000*y^16)/t^16].

Which answers describe the end behaviors of the function modeled by the graph? f(x)=−3x+2−2

Select each correct answer.

As x increases without bound, f(x) approaches the line y=−2.

As x increases without bound, f(x) increases without bound.

As x decreases without bound, f(x) approaches the line
y=−2.

As x increases without bound, f(x) decreases without bound.

Answers

For the given function, as x increases without bound, f(x) decreases without bound. Option D is correct.

Given that,
To describe the end behaviors of the function modeled by the graph,
f(x)=−3x+2−2

What are functions?

Functions is the relationship between sets of values. e g y=f(x), for every value of x there is its exists in set of y. x is the independent variable while Y is the dependent variable.

Here,
The function given is,
f(x) = -3x + 2 - 2
Given function contains equality so the function has no boundaries and the function is described from negative infinity to positive infinity, But the function has inverse behavior as when x is negative the function is positive and when x is positive function is negative.

Thus, for the given function, as x increases without bound, f(x) decreases without bound. Option D is correct.

Learn more about function here:

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Answer:

As x increases without bound, f(x) decreases without bound.

What is the general form of the equation of a circle with center at (a, b) and radius of length m?. A. x2 + y2 − 2ax − 2by + (a2 + b2 − m2) = 0 . . B.x2 + y2 + 2ax + 2by + (a2 + b2 − m2) = 0. C. x2 + y2 − 2ax − 2by + (a + b − m2) = 0. D. x2 + y2 + 2ax + 2by + a2 + b2 = -m2.

Answers

The generic equation of the circle is given by:
 (x-xo) ^ 2 + (y-yo) ^ 2 = r ^ 2
 Where,
 r: radius of the circle
 (xo, yo): center of the circle.
 The center of the circle in this case is:
 (xo, yo): (a, b)
 The radius of the circle is:
 r = m
 Substituting values we have:
 (x-a) ^ 2 + (y-b) ^ 2 = m ^ 2
 Rewriting the equation we have:
 x ^ 2 - 2ax + a ^ 2 + y ^ 2 - 2by + b ^ 2 - m ^ 2 = 0x ^ 2 + y ^ 2 - 2ax - 2by + a ^ 2 + b ^ 2 - m ^ 2 = 0
 Answer:
 The general form of the equation of a circle with center at (a, b) and radius of length m is:
 x ^ 2 + y ^ 2 - 2ax - 2by + a ^ 2 + b ^ 2 - m ^ 2 = 0
 option A
The general form of the equation of a circle with center at (a, b) and radius of length m is B.x2 + y2 + 2ax + 2by + (a2 + b2 − m2)


the standard equation for a circle is: (x-a)^2 + (y-b)^2 = m^2

x-a)^2 + (y-b)^2 = m^2 

means(x-a)(x-a) + (y-b)(y-b) = m^2 then you foil

x^2 - 2ax + a^2 + y^2 - 2by + b^2 = m^2

What is the equation for a line that passes through the points (-5,-4) and (10,14)?

Answers

Answer:

y=6/5x+2

Step-by-step explanation:

1.find the slope(m)

m=y2-y1/x2-x1

m=14-(-4)/10-(-5)

m=14+4/10+5

m=18/15

m=6/5

y-y1=m(x-x1)

y-(-4)=6/5(x-(-5))

y+4=6/5(x+5)

y+4=6/5x+30/5

y=6/5x+30/5-4

y=6/5x+10/5

y=6/5x+2
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